Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
  ↳18 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)

The TRS R 2 is

a → b
a → c

The signature Sigma is {c, a, b}

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROD(xs) → PRODITER(xs, s(0))
PRODITER(xs, x) → IFPROD(isempty(xs), xs, x)
PRODITER(xs, x) → ISEMPTY(xs)
IFPROD(false, xs, x) → PRODITER(tail(xs), times(x, head(xs)))
IFPROD(false, xs, x) → TAIL(xs)
IFPROD(false, xs, x) → TIMES(x, head(xs))
IFPROD(false, xs, x) → HEAD(xs)
PLUS(s(x), y) → PLUS(x, y)
TIMES(x, y) → TIMESITER(x, y, 0, 0)
TIMESITER(x, y, z, u) → IFTIMES(ge(u, x), x, y, z, u)
TIMESITER(x, y, z, u) → GE(u, x)
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, plus(y, z), s(u))
IFTIMES(false, x, y, z, u) → PLUS(y, z)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


GE(s(x), s(y)) → GE(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GE(x1, x2)  =  GE(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
s1: [1]
GE1: [1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[PLUS2, s1]

Status:
PLUS2: [2,1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

We have to consider all minimal (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IFTIMES(false, x, y, z, u) → TIMESITER(x, y, plus(y, z), s(u))
TIMESITER(x, y, z, u) → IFTIMES(ge(u, x), x, y, z, u)

The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

We have to consider all minimal (P,Q,R)-chains.

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IFPROD(false, xs, x) → PRODITER(tail(xs), times(x, head(xs)))
PRODITER(xs, x) → IFPROD(isempty(xs), xs, x)

The TRS R consists of the following rules:

prod(xs) → prodIter(xs, s(0))
prodIter(xs, x) → ifProd(isempty(xs), xs, x)
ifProd(true, xs, x) → x
ifProd(false, xs, x) → prodIter(tail(xs), times(x, head(xs)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0, 0)
timesIter(x, y, z, u) → ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, plus(y, z), s(u))
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → b
a → c

The set Q consists of the following terms:

prod(x0)
prodIter(x0, x1)
ifProd(true, x0, x1)
ifProd(false, x0, x1)
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2, x3)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
a

We have to consider all minimal (P,Q,R)-chains.